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Elanso [62]
3 years ago
13

Please help, I’ll mark your answer as brainliest!

Mathematics
1 answer:
eimsori [14]3 years ago
7 0
The answer would be
7b+4
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Help me..................
Andrew [12]

Answer:

while looking ata number for example 1234, 1 is the thousands place 2 is the hundreds place 3 is the tenths place and 4 is the ones place

Step-by-step explanation:

1 is the thousands place because 1234 is a number consisting of 1000 so 1 in 1234 is the thousnads place and if we keep going youll see that its pretty easy from there

5 0
3 years ago
What number is 60 percent of 21.8
Mrac [35]
So, first, since 21.8 is a whole thing and you need a part, write x/21.8. Then, compare it to 60/100, or 60%. 100/21.8=4.76. So, divide 60/4.76 and your answer is 12.6.

Glad to help.

-A
8 0
4 years ago
Please help me solve this
damaskus [11]
N=8
x= -19

Let me know if you need to see how to solve it
8 0
3 years ago
Alonzo works mowing lawns and babysitting. He earns $8.20 an hour for mowing and $7.90 an hour for babysitting.How much will he
Y_Kistochka [10]

Answer:

Alonzo would earn $65.3

Step-by-step explanation:

8.2*7=57.4

7.9*1=7.9

57.4+7.9=65.3

6 0
3 years ago
A ½in diameter rod of 5in length is being considered as part of a mechanical linkagein which it can experience a tensile loading
Evgesh-ka [11]

Answer:

a. Maximum Load = Force = 27085.09 N

b. Maximum Energy = 3.440 Joules

Step-by-step explanation:

Given

Rod diameter = ½in = 0.5in

Length = 5in

Young's modulus = 15.5Msi

By applying the 0.2% offset rule,

The maximum load the rod can hold before it gets to breaking point is given as follows by taking the strain as 0.2%

Young Modulus = Stress/Strain ------- Make Stress the Subject of Formula

Stress = Strain * Young Modulus

Stress = 0.2% * 15.5

Stress = 0.002 * 15.5

Stress = 0.031Msi

Calculating the area of the rod

Area = πr² or πd²/4

Area, A = 22/7 * 0.5^4 / 4

A = 22/7 * 0.25 / 4

A = 5.5/28

A = 0.1964in²

The maximum load that the rod would take before it starts to permanently elongate is given by

Force = Stress * Atea

Force = 0.031Msi * 0.1964in²

Force = 31Ksi * 0.1964in²

Force = 6.089Ksi in²

Force = 6.089 * 1000lbf

Force = 6089 lbf

1 lbf = 4.4482N

So, Force = 6089 * 4.4482N

Force = 27085.09 N

b.

Using Strain to Energy Formula

U = V×σ²/2·E

Where V = Volume

V = Length * Area

V = 5 in * 0.1964in²

V = 0.982in³

σ = Stress = 0.031Msi

= 0.031 * 1000Ksi

= 31Ksi

= 31 * 1000psi

= 31000psi

E = Young Modulus = 15.5Msi

= 15.5 * 1000Ksi

= 15.5 * 1000 * 1000psi

= 15500000psi

So,

Energy = 0.982 * 31000²/ ( 2 * 15500000)

Energy = 943,702,000/31000000

Energy = 30.442in³psi

------- Converted to ftlbf

Energy = 2.537 ftlbf

-------- Converted to Joules

Energy = 3.440 Joules

7 0
4 years ago
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